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Chapter 10

Query Processing: The Basics

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External Sorting

• Sorting is used in implementing many relational operations

• Problem: – Relations are typically large, do not fit in main memory

– So cannot use traditional in-memory sorting algorithms

• Approach used:– Combine in-memory sorting with clever techniques aimed at

minimizing I/O

– I/O costs dominate => cost of sorting algorithm is measured in the number of page transfers

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External Sorting (cont’d)

• External sorting has two main components:– Computation involved in sorting records in

buffers in main memory– I/O necessary to move records between mass

store and main memory

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Simple Sort Algorithm• M = number of main memory page buffers• F = number of pages in file to be sorted• Typical algorithm has two phases:

– Partial sort phase: sort M pages at a time; create F/M sorted runsruns on mass store, cost = 2F

Example: M = 2, F = 7run

Original file

Partially sorted file

5 3 2 6 1 10 15 7 20 11 8 4 7 5

2 3 5 6 1 7 10 15 4 8 11 20 5 7

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Simple Sort Algorithm– Merge Phase: merge all runs into a single run

using M-1 buffers for input and 1 output buffer• Merge step: divide runs into groups of size M-1 and merge each group into a run; cost = 2F

each step reduces number of runs by a factor of M-1

M pagesBuffer

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Merge: An Example

2 3 5 6

1 7 10 15

Input buffersOutput buffer

1 2 3 5 6 7 10 15

2 3

1 7

5 6

10 15

1 23 56 710 15

Output runInput runs

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Simple Sort Algorithm

• Cost of merge phase: – (F/M)/(M-1)k runs after k merge steps Log M-1(F/M) merge steps needed to merge an

initial set of F/M sorted runs

– cost = 2F Log M-1(F/M) 2F(Log M-1F -1)

• Total cost = cost of partial sort phase + cost of merge phase 2F Log M-1F

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Duplicate Elimination

• A major step in computing projection, union, and difference relational operators

• Algorithm:– Sort– At the last stage of the merge step eliminate

duplicates on the fly– No additional cost (with respect to sorting) in

terms of I/O

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Duplicate elimination During Merge

2 3 5 6

1 3 5 15

Input buffersOutput buffer

1 2 3 5 6 15

2 3

1 3

5 6

5 15

1 23 56 15

Output runInput runs Last key used

1215356

Key 3 ignored: duplicate

Key 5 ignored: duplicate

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Sort-Based Projection

• Algorithm:– Sort rows of relation at cost of 2F Log M-1F

– Eliminate unwanted columns in partial sort phase (no additional cost)

– Eliminate duplicates on completion of last merge step (no additional cost)

• Cost: the cost of sorting

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Hash-Based Projection• Phase 1:

– Input rows– Project out columns– Hash remaining columns using a hash function with range 1…M-1

creating M-1 buckets on disk– Cost = 2F

• Phase 2: – Sort each bucket to eliminate duplicates– Cost (assuming a bucket fits in M-1 buffer pages) = 2F

• Total cost = 4F

M pages

Buffer

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Computing Selection (attr op value)

• No index on attr:– If rows are not sorted on attr:

• Scan all data pages to find rows satisfying selection condition

• Cost = F

– If rows are sorted on attr and op is =, >, < then: • Use binary search (at log2 F ) to locate first data

page containing row in which (attr = value)

• Scan further to get all rows satisfying (attr op value)

• Cost = log2 F + (cost of scan)

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Computing Selection (attr op value)

• Clustered B+ tree index on attr (for “=” or range search):

– Locate first index entry corresponding to a row in which (attr = value). CostCost = depth of tree

– Rows satisfying condition packed in sequence in successive data pages; scan those pages.

CostCost: number of pages occupied by qualifying rows

B+ treeindex entries(containing rows)that satisfycondition

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Computing Selection (attr op value)

• Unclustered B+ tree index on attr (for “=” or range search):

– Locate first index entry corresponding to a row in which (attr = value).

CostCost = depth of tree

– Index entries with pointers to rows satisfying condition are packed in sequence in successive index pages

• Scan entries and sort record Ids to identify table data pages with qualifying rows

Any page that has at least one such row must be fetched once.

• CostCost: number of rows that satisfy selection condition

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Unclustered B+ Tree Index

index entries (containingrow Ids) that satisfycondition

data page

Data file

B+ Tree

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Computing Selection (attr = value)

• Hash index on attr (for “=” search only):– Hash on value. CostCost 1.2

• 1.2 – typical average cost of hashing (> 1 due to possible overflow chains)

• Finds the (unique) bucket containing all index entries satisfying selection condition

• Clustered index – all qualifying rows packed in the bucket (a few pages)

CostCost: number of pages occupies by the bucket

• Unclustered index – sort row Ids in the index entries to identify data pages with qualifying rows

Each page containing at least one such row must be fetched once

CostCost: min(number of qualifying rows in bucket, number of pages in file)

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Computing Selection (attr = value)

• Unclustered hash index on attr (for equality search)

buckets

data pages

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Access Path• Access pathAccess path is the notion that denotes algorithm +

data structure used to locate rows satisfying some condition

• Examples:– File scan: can be used for any condition– Hash: equality search; all search key attributes of hash

index are specified in condition– B+ tree: equality or range search; a prefix of the search

key attributes are specified in condition• B+ tree supports a variety of access paths

– Binary search: Relation sorted on a sequence of attributes and some prefix of that sequence is specified in condition

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Access Paths Supported by B+ tree

• Example: Given a B+ tree whose search key is the sequence of attributes a2, a1, a3, a4

– Access path for search a1>5 AND a2=3 AND a3=‘x’ (R): find first entry having a2=3 AND a1>5 AND a3=‘x’ and scan leaves from there until entry having a2>3 or a3 ‘x’. Select satisfying entries

– Access path for search a2=3 AND a3 >‘x’ (R): locate first entry having a2=3 and scan leaves until entry having a2>3. Select satisfying entries

– Access path for search a1>5 AND a3 =‘x’ (R): Scan of R

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Choosing an Access Path• SelectivitySelectivity of an access path = number of pages

retrieved using that path• If several access paths support a query, DBMS

chooses the one with lowest selectivity• Size of domain of attribute is an indicator of the

selectivity of search conditions that involve that attribute

• Example: CrsCode=‘CS305’ AND Grade=‘B’ (TranscriptTranscript)

– a B+ tree with search key CrsCode has lower selectivity than a B+ tree with search key Grade

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Computing Joins• The cost of joining two relations makes the

choice of a join algorithm crucial• SimpleSimple block-nested loops block-nested loops join algorithm

for computing r A=B s

foreach page pr in r do foreach page ps in s do output pr A=B ps

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Block-Nested Loops Join

• If r and s are the number of pages in r and s, the cost of algorithm is

r + r s + cost of outputting final result

– If r and s have 103 pages each,

cost is 103 + 103 * 103

– Choose smaller relation for the outer loop:• If r < s then r + r s < s + r s

Number of scans of relation s

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Block-Nested Loops Join• Cost can be reduced to

r + (r/(M-2)) s + cost of outputting final result

by using M buffer pages instead of 1.

Number of scans of relation s

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Block-Nested Loop Illustrated

Output buffer

s

r

Input buffer for s

Input buffer for r

… and so on

r s

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Index-Nested Loop Join r A=B s

• Use an index on s with search key B (instead of scanning s) to find rows of s that match tr

– Cost Cost = r + r + cost of outputting final result

– Effective if number of rows of s that match tuples in r is small (i.e., is small) and index is clustered

foreach tuple tr in r do { use index to find all tuples ts in s satisfying tr.A=ts.B; output (tr, ts) }

Number of rows in r

avg cost of retrieving all rows in s that match tr

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Sort-Merge Join r A=B s

sort r on A;sort s on B;while !eof(r) and !eof(s) do { Scan r and s concurrently until tr.A=ts.B=c; Output A=c(r)B=c (s) }

r

s

B=c (s)

A=c(r)

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Join During Merge Illustrated

1 3p p

r

s

DA

BE

p p4 0

0 9q q

r9

8 7 3s s s

s7

t t2 5

u u u2 5 0

5 7u u

1 1v v

x0

1 3 1 3p p p pp p p p4 0 0 4

8 7 3s s ss s s7 7 7

5 7 5 7 5 7u u u u u uu u u u u u2 2 5 5 0 0

r A=B s

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Cost of Sort-Merge Join• CostCost of sorting assuming M buffers: 2 r log M-1 r + 2 s log M-1 s

• CostCost of merging:– Scanning A=c(r) and B=c (s) can be combined with the last step of

sorting of r and s --- costs nothing– Cost of A=c(r)B=c (s) depends on whether A=c(r) can fit in the

buffer• If yes, this step costs 0 • In no, each A=c(r)B=c (s) is computed using block-nested join, so the

cost is the cost of the join. (Think why indexed methods or sort-merge are inapplicable to Cartesian product.)

• Cost of outputting the final result depends on the size of the result

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Hash-Join r A=B s

• Step 1: Hash r on A and s on B into the same set of buckets

• Step 2: Since matching tuples must be in same bucket, read each bucket in turn and output the result of the join

• CostCost:: 3 (r + s ) + cost of output of final result– assuming each bucket fits in memory

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Hash Join

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Star Joins

• r cond1 r1 cond2

… condn rn

– Each cond i involves only the attributes of ri and r

r

r1

r2

r3

r4

r5

cond1 cond2

cond3

cond4

cond5

Star relationSatellite

relations

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Star Join

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Computing Star Joins

• Use join index join index (Chapter 11)– Scan r and the join index {<r,r1,…,rn>} (which is

a set of tuples of rids) in one scan

– Retrieve matching tuples in r1,…,rn

– Output result

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Computing Star Joins• Use bitmap indicesbitmap indices (Chapter 11)

– Use one bitmapped join index, Ji , per each partial join

r condi ri

– Recall: Ji is a set of <v, bitmap>, where v is an rid of a tuple in ri and bitmap has 1 in k-th position iff k-th tuple of r joins with the tuple pointed to by v

1. Scan Ji and logically OR all bitmaps. We get all rids in r that join with ri

2. Now logically AND the resulting bitmaps for J1, …, Jn.3. Result: a subset of r, which contains all tuples that can

possibly be in the star join • Rationale: only a few such tuples survive, so can use indexed loops

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Choosing Indices

• DBMSs may allow user to specify – Type (hash, B+ tree) and search key of index– Whether or not it should be clustered

• Using information about the frequency and type of queries and size of tables, designer can use cost estimates to choose appropriate indices

• Several commercial systems have tools that suggest indices– Simplifies job, but index suggestions must be verified

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Choosing Indices – Example

• If a frequently executed query that involves selection or a join and has a large result set, use a clustered B+ tree index

Example: Retrieve all rows of TranscriptTranscript for StudId

• If a frequently executed query is an equality search and has a small result set, an unclustered hash index is best– Since only one clustered index on a table is possible,

choosing unclustered allows a different index to be clustered

Example: Retrieve all rows of TranscriptTranscript for (StudId, CrsCode)